Дано: А(-2;4), В(3;3), C(5;-1) Найти:вектора AB; AC; |AB|; |AC| Cos<BAC

Решите задачу:

$\bar A\bar B=\left\{3-(-2);\;3-4\right\}=\{5;\;-1\}\\\bar A\bar C=\left\{5-(-2);\;-1-4\right\}=\{7;\;-5\}\\\left|AB\right|=\sqrt{5^2+(-1)^2}=\sqrt{25+1}=\sqrt{26}\\\left|AC\right|=\sqrt{7^2+(-5)^2}=\sqrt{49+25}=\sqrt{74}\\\cos\angle BAC=\frac{(\bar{AB},\bar{AC})}{|AB||AC|}=\frac{5\cdot7+(-1)\cdot(-5)}{\sqrt{26}\cdot\sqrt{74}}=\frac{35+5}{\sqrt{1924}}=\frac{40}{\sqrt{1924}}$